The question is: Let $\chi_1,\chi_2,\chi_3,\chi_4$ be Dirichlet characters in $(\mathbb Z/N\mathbb Z)^*$ and they are distinct pairwise.
Can we proof that $\chi_1+\chi_2\neq\chi_3+\chi_4$? Or can we give an example of $\chi_1+\chi_2=\chi_3+\chi_4$?
$~$
Background: I get this question when calculating weight $1$-Eisenstein series.
$~$
My approach: I can prove $\chi_1+\chi_2\neq\chi_3+\chi_4$ when $N$ is a power of prime. However, I don't know how to get a general result. The calculation as follows.
First assume that $N$ is a power of an odd prime, then $(\mathbb Z/N\mathbb Z)^*$ has a generator, say $r$. Then we can write all characters of $(\mathbb Z/N\mathbb Z)^*$ in the following form:
$$\chi_k(r^n)=e^{2\pi ikn/\varphi(N)},\ k=1,\cdots,\varphi(N).$$
Assuming $\chi_{k_1}+\chi_{k_2}=\chi_{k_3}+\chi_{k_4}$, then $\chi_{k_1}(r)+\chi_{k_2}(r)=\chi_{k_3}(r)+\chi_{k_4}(r)$. Since $\chi_{k_i}(r)$ are distinct pairwise, we obtain $\chi_{k_1}(r)+\chi_{k_2}(r)=\chi_{k_3}(r)+\chi_{k_4}(r)=0$. Thus $k_1-k_2\equiv\varphi(N)/2\ (\text{mod}\ \varphi(N))$. Moreover, we have $\chi_{k_1}(r^2)=\chi_{k_2}(r^2)$ and $\chi_{k_3}(r^2)=\chi_{k_4}(r^2)$. So by the assumption we have $\chi_{k_1}(r^2)=\chi_{k_3}(r^2)$, which yields $k_1-k_3\equiv\varphi(N)/2\ (\text{mod}\ \varphi(N))$ and hence $k_2=k_3$, a contradiction.
Next we assume that $N=2^t\ (t\geq3)$ (Since we don't have $4$ distinct characters in the cases $N=2,4$). It is known that $(\mathbb Z/N\mathbb Z)^*=\langle-1,5\rangle$ and the element $5$ generates a subgroup of order $2^{t-2}$. In this case, we know the expressions of all characters, too. Let $\chi_{k,m}\ (k=1,\cdots,2^{t-2};m=1,2)$ denotes all characters, where
$$\chi_{k,m}(5)=e^{2\pi ik/2^{t-2}}\text{ and }\chi_{k,m}(-1)=(-1)^m.$$
Assuming $\chi_{k_1,m_1}+\chi_{k_2,m_2}=\chi_{k_3,m_3}+\chi_{k_4,m_4}$, then $\chi_{k_1,m_1}(5)+\chi_{k_2,m_2}(5)=\chi_{k_3,m_3}(5)+\chi_{k_4,m_4}(5)$. Moreover, $\chi_{k,m}(5)$ repeats exactly twice (When $k_1=k_2$ but $m_1\neq m_2$).
- If $\chi_{k_1,m_1}(5)=\chi_{k_3,m_3}(5)$ and $\chi_{k_2,m_2}(5)=\chi_{k_4,m_4}(5)$, without loss of generality, we have $\chi_{k_1,m_1}(-1)=1$ and $\chi_{k_3,m_3}(-1)=-1$. Now we consider
$$\chi_{k_1,m_1}(-1)+\chi_{k_2,m_2}(-1)=\chi_{k_3,m_3}(-1)+\chi_{k_4,m_4}(-1).$$
Thus $\chi_{k_2,m_2}(-1)=-1$ and $\chi_{k_4,m_4}(-1)=1$. Thereby from
$$\chi_{k_1,m_1}(-5)+\chi_{k_2,m_2}(-5)=\chi_{k_3,m_3}(-5)+\chi_{k_4,m_4}(-5)$$
we obtain $\chi_{k_1,m_1}(5)-\chi_{k_2,m_2}(5)=-\chi_{k_3,m_3}(5)+\chi_{k_4,m_4}(5)$, which implies $\chi_{k_1,m_1}(5)=\chi_{k_4,m_4}(5)$, a contradiction.
- If $\chi_{k,m}(5)$ are distinct pairwise, then $\chi_{k_1,m_1}(5)+\chi_{k_2,m_2}(5)=\chi_{k_3,m_3}(5)+\chi_{k_4,m_4}(5)=0$. Similarly we get $k_2=k_3$ and so $\chi_{k_2,m_2}(5)=\chi_{k_3,m_3}(5)$, which contradicts the assumption.
Any comment is appreciated :)
